Chapter 5, Problem 16A

Water balloons of different masses are launched by slingshots at different launching velocities v. All have the same vertical component of launching velocities.

a. Rank by the time in the air, from longest to shortest.

b. Rank by the maximum height reached, from highest to lowest.

c. Rank by the maximum range, from greatest to least.

To determine

To rank: The balloons on the basis of time in the air, from longest to shortest.

Answer to Problem 16A

The rank of balloons is, $A=B=C=D$

Explanation of Solution

Given:

Water balloons of different masses are launched by slingshots at different launching velocities as shown below.

Formula used:

Horizontal component of initial velocity of projectile is,

  ${\text{u}}_{\text{x}}\text{ = ucos}\left(\text{θ}\right)$

Vertical component of initial velocity of projectile is

  ${\text{u}}_{\text{y}}\text{ = usin}\left(\text{θ}\right)$

Where,

u is the initial velocity.

Time of flight is the amount of time required for projectile to complete its trajectory,

  $\text{T}\text{=}\frac{\text{2usin}\left(\text{θ}\right)}{\text{g}}$

Calculation:

Consider case ( A )

Vertical component of velocity is,

  $\begin{array}{c}{\text{u}}_{\text{y}}\text{ = usin}\left(\text{θ}\right)\\ \text{u}\text{=}\frac{{\text{u}}_{\text{y}}}{\text{sin}\left(\text{θ}\right)}\\ \text{=}\frac{\text{20}\text{m/s}}{\text{sin}\left(\text{37}°\right)}\\ =\text{33}\text{.23}\text{m/s}\end{array}$

Time of flight is

  $\begin{array}{c}\text{T}\text{=}\frac{\text{2usin}\left(\text{θ}\right)}{\text{g}}\\ \text{=}\frac{\text{2}\left(\text{33}\text{.23}\text{m/s}\right)\text{sin}\left({\text{37}}^{\text{0}}\right)}{\text{9}\text{.8}{\text{m/s}}^{\text{2}}}\\ \text{T}\text{=}\text{4}\text{.081}\text{sec}\end{array}$

Consider case ( B )

Vertical component of velocity is,

  $\begin{array}{c}{\text{u}}_{\text{y}}\text{ = usin}\left(\text{θ}\right)\\ \text{u}\text{=}\frac{{\text{u}}_{\text{y}}}{\text{sin}\left(45\right)}\\ \text{=}\frac{\text{20}\text{m/s}}{\text{sin}\left({45}^{\text{0}}\right)}\\ \text{=}28.28\text{m/s}\end{array}$

Time of flight is

  $\begin{array}{c}\text{T}\text{=}\frac{\text{2usin}\left(\text{θ}\right)}{\text{g}}\\ \text{=}\frac{\text{2}\left(\text{28}\text{.28}\text{m/s}\right)\text{sin}\left({45}^{\text{0}}\right)}{\text{9}\text{.8}{\text{m/s}}^{\text{2}}}\\ \text{T}\text{=}\text{4}\text{.081}\text{sec}\end{array}$

Consider case ( C )

Vertical component of velocity is,

  $\begin{array}{c}{\text{u}}_{\text{y}}\text{ = usin}\left(\text{θ}\right)\\ \text{u}\text{=}\frac{{\text{u}}_{\text{y}}}{\text{sin}\left(\text{θ}\right)}\\ \text{=}\frac{\text{20}\text{m/s}}{\text{sin}\left({\text{37}}^{\text{0}}\right)}\\ =\text{33}\text{.23}\text{m/s}\end{array}$

Time of flight is

  $\begin{array}{c}\text{T}\text{=}\frac{\text{2usin}\left(\text{θ}\right)}{\text{g}}\\ \text{=}\frac{\text{2}\left(\text{33}\text{.23}\text{m/s}\right)\text{sin}\left({\text{37}}^{\text{0}}\right)}{\text{9}\text{.8}{\text{m/s}}^{\text{2}}}\\ \text{=}\text{4}\text{.081}\text{sec}\end{array}$

Consider case ( D )

Vertical component of velocity is,

  $\begin{array}{c}{\text{u}}_{\text{y}}\text{ = usin}\left(\text{θ}\right)\\ \text{u}\text{=}\frac{{\text{u}}_{\text{y}}}{\text{sin}\left(45\right)}\\ \text{=}\frac{\text{20}\text{m/s}}{\text{sin}\left({45}^{\text{0}}\right)}\\ \text{=}28.28\text{m/s}\end{array}$

Time of flight is

  $\begin{array}{c}\text{T}\text{=}\frac{\text{2usin}\left(\text{θ}\right)}{\text{g}}\\ \text{=}\frac{\text{2}\left(\text{28}\text{.28}\text{m/s}\right)\text{sin}\left({45}^{\text{0}}\right)}{\text{9}\text{.8}{\text{m/s}}^{\text{2}}}\\ \therefore \text{T}\text{=}\text{4}\text{.081}\text{sec}\end{array}$

Conclusion:

Therefore, time in the air, from longest to shortest is, $A=B=C=D$

To determine

To rank: The balloons on the basis of maximum height reached, from highest to lowest.

Answer to Problem 16A

The rank of balloons is, $A=B=C=D$

Explanation of Solution

Given:

Water balloons of different masses are launched by slingshots at different launching velocities as shown below.

Formula used:

  $\begin{array}{l}\text{Maximum height at which particle rached in the vertical direction is}\\ \therefore \text{H}\text{=}\frac{{\text{u}}^{\text{2}}{\text{sin}}^{\text{2}}\left(\text{θ}\right)}{\text{2g}}\\ \text{Where,}\text{g=}\text{acceleration}\text{due}\text{to}\text{gravity}\text{=}\text{9}\text{.8}{\text{m/s}}^{\text{2}}\end{array}$

Calculation:

Consider case ( A )

Maximum height is,

  $\begin{array}{l}\text{H}\text{=}\frac{{\text{u}}^{\text{2}}{\text{sin}}^{\text{2}}\left(\text{θ}\right)}{\text{2g}}\\ \text{=}\frac{{\left(\text{33}\text{.23}\text{m/s}\right)}^{\text{2}}{\text{sin}}^{\text{2}}\left({\text{37}}^{\text{0}}\right)}{\text{2}\left(\text{9}\text{.8}{\text{m/s}}^{\text{2}}\right)}\\ \text{H=}\text{20}\text{.40}\text{m}\end{array}$

Consider case ( B )

  $\begin{array}{l}\text{Maximum height is,}\\ \text{H}\text{=}\frac{{\text{u}}^{\text{2}}{\text{sin}}^{\text{2}}\left(\text{θ}\right)}{\text{2g}}\\ \text{=}\frac{{\left(\text{28}\text{.28}\text{m/s}\right)}^{\text{2}}{\text{sin}}^{\text{2}}\left({\text{45}}^{\text{0}}\right)}{\text{2}\left(\text{9}\text{.8}{\text{m/s}}^{\text{2}}\right)}\\ \text{H=}\text{20}\text{.40}\text{m}\end{array}$

Consider case ( C )

  $\begin{array}{l}\text{Maximum height is,}\\ \text{H}\text{=}\frac{{\text{u}}^{\text{2}}{\text{sin}}^{\text{2}}\left(\text{θ}\right)}{\text{2g}}\\ \text{=}\frac{{\left(\text{33}\text{.23}\text{m/s}\right)}^{\text{2}}{\text{sin}}^{\text{2}}\left({\text{37}}^{\text{0}}\right)}{\text{2}\left(\text{9}\text{.8}{\text{m/s}}^{\text{2}}\right)}\\ \text{H=}\text{20}\text{.40}\text{m}\end{array}$

Consider case ( D )

  $\begin{array}{l}\text{Maximum height is,}\\ \text{H}\text{=}\frac{{\text{u}}^{\text{2}}{\text{sin}}^{\text{2}}\left(\text{θ}\right)}{\text{2g}}\\ \text{=}\frac{{\left(\text{33}\text{.23}\text{m/s}\right)}^{\text{2}}{\text{sin}}^{\text{2}}\left({\text{37}}^{\text{0}}\right)}{\text{2}\left(\text{9}\text{.8}{\text{m/s}}^{\text{2}}\right)}\\ \text{H=}\text{20}\text{.40}\text{m}\end{array}$

Conclusion:

Maximum height reached, from highest to lowest is, $A=B=C=D$

To determine

To rank: The balloons on the basis of maximum range, from greatest to least.

Answer to Problem 16A

Maximum range, from greatest to least is, $A=C>B=D$

Explanation of Solution

Given:

Water balloons of different masses are launched by slingshots at different launching velocities as shown below.

Formula used:

Range of projectile is,

  $R=\frac{u^2\sin \left(2\theta \right)}{g}$

Calculation:

Consider case ( A )

Range of projectile is,

  $\begin{array}{l}R=\frac{u^2\sin \left(2\theta \right)}{g}\\ R=\frac{{\left(33.23\text{m/s}\right)}^2\sin \left(2\left({37}^0\right)\right)}{9.8{\text{m/s}}^{\text{2}}}\\ R\text{=}\text{108}\text{.31}\text{m}\end{array}$

Consider case ( B )

Range of projectile is

  $\begin{array}{l}R=\frac{u^2\sin \left(2\theta \right)}{g}\\ R=\frac{{\left(28.28\text{m/s}\right)}^2\sin \left(2\left({45}^0\right)\right)}{9.8{\text{m/s}}^{\text{2}}}\\ R=81.60\text{m}\end{array}$

Consider case ( C )

Range of projectile is,

  $\begin{array}{l}R=\frac{u^2\sin \left(2\theta \right)}{g}\\ R=\frac{{\left(33.23\text{m/s}\right)}^2\sin \left(2\left({37}^0\right)\right)}{9.8{\text{m/s}}^{\text{2}}}\\ R=108.31\text{m}\end{array}$

Consider case ( D )

Range of projectile is,

  $\begin{array}{l}R=\frac{u^2\sin \left(2\theta \right)}{g}\\ R=\frac{{\left(28.28\text{m/s}\right)}^2\sin \left(2\left({45}^0\right)\right)}{9.8{\text{m/s}}^{\text{2}}}\\ R=81.60\text{m}\end{array}$

Conclusion:

Therefore, maximum range, from greatest to least is, $A=C>B=D$